Answer:
40 minutes
Step-by-step explanation:
The circumference of the larger circle is ...
C = 2πr = 2π(5 in) = 10π in
The bug navigates the circumference at 3π in/min, so will take
time = distance/speed = (10π in)/(3π in/min) = 10/3 min
to travel once around.
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The circumference of the smaller circle is ...
C = 2πr = 2π(2 in) = 4π in
The bug navigates this circumference at 2.5π in/min, so will take
(4π in)/(2.5π in/min) = 8/5 min
to travel once around.
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The bugs will meet at a time that is the least common multiple of these times. Both can be expressed in 15ths of a minute as ...
{50/15, 24/15}
Then the LCM of these will be ...
(1/15)LCM(50, 24) = (1/15)(50×24)/GCD(50, 24) = 1200/30 = 40
It will be 40 minutes before the bugs next meet at point P.
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A graphing calculator can make use of the mod function to show when the bugs meet at point P (total is displacement of the two bugs from P is zero). It shows the meeting occurs after 40 minutes.
Answer:
40
Step-by-step explanation:
The circumference of the larger circle, , is . The circumference of the smaller circle, , is . The bug on crawls the circumference in minutes, while the bug on crawls the circumference in minutes. The two bugs will meet at point P in some minutes, when and are both integers. We have , so we have to find the LCM of and . The LCM is , so the bugs will next meet in minutes.
